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Sunday, July 4, 2010. A jar begins with one amoeba.Every minute , every amoeba turns into 0 ,1,2 or 3 with 25% probability for each case (dies, does nothing,splits into 2 , splits into 3).What is the probability that amoeba population eventually dies out? Saturday, March 20, 2010. 1) Linked List reversal , without using any extra memory. (Code Insertion). 2) M 0's N 1's M N. No. of possible permutations of String of length M N. 4) Merge two trees into a new tree. 5) Consider linked list {1,2,3,4,5,6}.

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Sunday, July 4, 2010. A jar begins with one amoeba.Every minute , every amoeba turns into 0 ,1,2 or 3 with 25% probability for each case (dies, does nothing,splits into 2 , splits into 3).What is the probability that amoeba population eventually dies out? Saturday, March 20, 2010. 1) Linked List reversal , without using any extra memory. (Code Insertion). 2) M 0's N 1's M N. No. of possible permutations of String of length M N. 4) Merge two trees into a new tree. 5) Consider linked list {1,2,3,4,5,6}.

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Finally MyBlog | saikacollection.blogspot.com Reviews

https://saikacollection.blogspot.com

Sunday, July 4, 2010. A jar begins with one amoeba.Every minute , every amoeba turns into 0 ,1,2 or 3 with 25% probability for each case (dies, does nothing,splits into 2 , splits into 3).What is the probability that amoeba population eventually dies out? Saturday, March 20, 2010. 1) Linked List reversal , without using any extra memory. (Code Insertion). 2) M 0's N 1's M N. No. of possible permutations of String of length M N. 4) Merge two trees into a new tree. 5) Consider linked list {1,2,3,4,5,6}.

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saikacollection.blogspot.com

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Finally MyBlog: Interview Questions: Flipkart

http://saikacollection.blogspot.com/2010/03/interview-questions-flipkart.html

Saturday, March 20, 2010. 1) Linked List reversal , without using any extra memory. (Code Insertion). 2) M 0's N 1's M N. No. of possible permutations of String of length M N. 3) Count no of characters/words/lines in a file. Given file handle definition. 4) Merge two trees into a new tree. Example: (1,(2,null) and (1,(null,3) results in (1,(2,3). 1,(2,null) and (1,(3,null) results in null (Not a valid operation). 5) Consider linked list {1,2,3,4,5,6}. For(int i = 0;i. 9) Monty Hall Problem Card variant.

2

Finally MyBlog: PJ Puzzle

http://saikacollection.blogspot.com/2010/02/pj-puzzle.html

Tuesday, February 23, 2010. In which direction is the bus moving? February 24, 2010 at 12:32 AM. Just a common sense. We often miss what are obvious. The answer always lies in the place where you look at the last. Note: Only a member of this blog may post a comment. Subscribe to: Post Comments (Atom). Blind Man and Cards. How will you measure? Adobe Written Test Section II. Adobe Written Test Section I. View my complete profile.

3

Finally MyBlog: Hungry Elephant

http://saikacollection.blogspot.com/2010/02/hungry-elephant.html

Thursday, February 25, 2010. This one is easy and most famous puzzle. But I have a doubt here which I would post later after getting answers . There are 2 cities A and B, distance between them is 1000 Km. We have 3000 bananas in city A and a elephant. Can carry max 1000 bananas at any given time and it needs to eat a banana every 1 Km. How Many Max. No. of bananas you can transfer to city B? March 2, 2010 at 6:23 PM. I feel at max you can take only one banana to B. April 4, 2010 at 6:42 PM.

4

Finally MyBlog: March 2010

http://saikacollection.blogspot.com/2010_03_01_archive.html

Saturday, March 20, 2010. 1) Linked List reversal , without using any extra memory. (Code Insertion). 2) M 0's N 1's M N. No. of possible permutations of String of length M N. 3) Count no of characters/words/lines in a file. Given file handle definition. 4) Merge two trees into a new tree. Example: (1,(2,null) and (1,(null,3) results in (1,(2,3). 1,(2,null) and (1,(3,null) results in null (Not a valid operation). 5) Consider linked list {1,2,3,4,5,6}. For(int i = 0;i. 9) Monty Hall Problem Card variant.

5

Finally MyBlog: Blind Man and Cards

http://saikacollection.blogspot.com/2010/02/blind-man-and-cards.html

Monday, February 22, 2010. Blind Man and Cards. A blind man is handed a deck of 52 cards and told that exactly 10 of these cards are facing up. How can he divide the cards into two piles, not necessarily of equal size, with each pile having the same number of cards facing up? February 22, 2010 at 6:16 PM. Divide cards into 2 groups one having 10 cards (group I) and other having 42 cards (group II). Now flip all the cards in group-I , so we will have 3 facing up . February 23, 2010 at 5:21 PM.

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